Infrared LED brightness - Page 2 - maemo.org

vi_ | # 11 | 2012-02-29, 13:58 | Report

According to schematics, the current through the LEDs is controlled by the the 220R resistors. Replace them all with 110R resistors for maximum IR lols.

...Although, the fact they have used two transistors in parallel implies that the forward current is already quite high for them (the transistors). So, will doubling it cause any problems? If not, why use 2 transistors?? Can the LED even take any MOAR juice anyway?

Who will try to find out?

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sixwheeledbeast | # 12 | 2012-02-29, 19:04 | Report

Originally Posted by vi_ View Post

According to schematics, the current through the LEDs is controlled by the the 220R resistors. Replace them all with 110R resistors for maximum IR lols.

...Although, the fact they have used two transistors in parallel implies that the forward current is already quite high for them (the transistors). So, will doubling it cause any problems? If not, why use 2 transistors?? Can the LED even take any MOAR juice anyway?

Who will try to find out?

AFAIK messing with transistors linked like this can cause problems.
Two transistors in parallel like that have to have the same resistance (with not much tolerance).
If there not one transistor will end up taking all of the load. This will destroy both transistors quickly.

Two transistors allow double the current of one.

Most 5mm 860nm infra-red LED's I have seen have a max forward voltage of 1.8V. 940nm LED's can have even lower voltage (1.2-1.6)
Increasing voltage above this will reduce light output due to increased temperature and eventually fry the LED.

I am certainly not going to try.

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Mike Fila | # 13 | 2012-02-29, 21:15 | Report

Originally Posted by Copernicus View Post

Actually, make it a transmitter and a receiver, and I could turn the N900 into a "learning" remote.

(And it might be useful for the N9 too, if it has USB host mode.)

I'm not a hardware guy, so I don't really do this sort of thing, but I think it'd be easy to rework the Pierogi back-end to tie it to other IR devices...

It fairly easy to make a receiver and cheap

http://www.networkedmediatank.com/sh....php?tid=29013

that is a receiver using a1/8 phono jack, Im sure you could use a mini usb wire instead. For a sender there are plenty that are pre made for iphones using the headphone jack or that could easily be made as well

blue_led | # 14 | 2012-02-29, 23:31 | Report

Originally Posted by sixwheeledbeast View Post

AFAIK messing with transistors linked like this can cause problems.
Two transistors in parallel like that have to have the same resistance (with not much tolerance).
If there not one transistor will end up taking all of the load. This will destroy both transistors quickly.

Two transistors allow double the current of one.

Most 5mm 860nm infra-red LED's I have seen have a max forward voltage of 1.8V. 940nm LED's can have even lower voltage (1.2-1.6)
Increasing voltage above this will reduce light output due to increased temperature and eventually fry the LED.

I am certainly not going to try.

you are wrong
case ONLY one transistor is on the current flow thru this transistor ( one , singleton ) will be the same as both transistors are on state. only the led current will double wen both transistors are on.
Ic( max) = ( Vbat - Vled - Vcesat ) / ( R / 2 ) -----> ~ 15 mA
all values can be considered constants !
i am sure, transistor can sink more than that

the maximum brightness without mods is achieved on 1/2 duty cycle. ( register settings )

average led current will be ( 50% duty , both on ) : 1/2 * 15 mA * 2 = 15 mA and this value can be supported by any led continuous mode
!
AFIK peak led current on traditional ir remote is 1A !!! much higher than n900 ir led current
across all n900 the ir power is low compared with an genuine ir remote control

Last edited by blue_led; 2012-03-01 at 01:15.

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sixwheeledbeast | # 15 | 2012-03-01, 17:41 | Report

Originally Posted by blue_led View Post

you are wrong

This confirms what I thought.

Originally Posted by

On to paralleling the transistors - yes, you can parallel them, but it isn't as simple as just attaching the leads together. Transistors experience a positive temperature coefficient in gain. So higher temperatures mean less voltage drop. With two in parallel, they won't be identical to start with so one will have a little less voltage drop, which means more current flows through it (current follows the path of least resistance). So more heat, which raises the temperature of that transistor, causing a greater imbalance, until that transistor is hogging enough current to overheat it and burn it up. Then the second transistor has all the load and also burns up.

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blue_led | # 16 | 2012-03-01, 20:37 | Report

@ sixwheeledbeast
you have the right to believe what you want but who wants to be always right lose the chance to know the truth

yahoo answers in not a reference, at least for me, over my personal achievement when i designed animated led panels for public advertising . ( from processor assembly code to schematics and mechanical design )

finally, who is right ? you or nokia engineers ?
over and out

sixwheeledbeast | # 17 | 2012-03-01, 21:19 | Report

@blue_led
It wasn't the only reference I found, but the best explained one.
This is what I was taught, not what I believe.

It sounds like you have a lot more experience in this, so naturally I will take what you say as the truth.

Finally, why does this have anything to do with comparing myself and Nokia or there engineers?

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davdav | # 18 | 2012-03-02, 08:58 | Report

so now we must ask H-E-N developers to attach an external ir device ?

dani846 | # 19 | 2012-03-08, 17:30 | Report

still nothing for me?very sad.

woody14619 | # 20 | 2012-03-08, 19:04 | Report

Just a few things:

The brightness of the IR transmitter in the N900 is what it is. The best you can do is select the 1/2 duty cycle. You can't make it brighter with software, nor would you want to if you could.

As for the bits about transistors: The case sixwheeledbeast is making is valid for some applications, but not this one. Part of the reason being, the system is controlling the duty cycle on the led, so the chances of either heating up enough to cause this effect is minimal.

As for why it's using two, it is probably to control current. When one is on, there are only 2 resistors (equaling 110 miliohm), when both are on, there are 4 (equaling 55 miliohm). And yes, the calculations involved are simple enough to determine that most leds would have no problem handling that.

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